# The Direct Product of a finite number of cyclic groups, is again a cyclic group?

Can anyone please brief me on this question, do you prove or disprove? If it is to be disproved can you give a counterexample? if there is no generator, then what is an example of such a Direct Product of a finite number of cyclic groups?

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It is false, for example take $\mathbb{Z}_2^2$ it's a group of order 4, but all the non identity elements have order 2 –  Daniel Feb 11 '13 at 3:02
thanks for you answer –  Faye Feb 11 '13 at 3:05
You may find information on the Chinese Remainder Theorem helpful (in the language of abstract algebra rather than elementary number theory); in particular its failure when its hypotheses are not met. Also helpful: the fundamental theorem of abelian groups. –  anon Feb 11 '13 at 3:09

You have to play a little bit in the playground of groups. Searching for the smallest example: The smallest non-trivial cyclic group is $\mathbb Z_2$. So, why not try $\mathbb Z_2\times \mathbb Z_2$? Do you know how many elements are there in $\mathbb Z_2\times \mathbb Z_2$? (not too many!). Try each one of them and see if it generates the entire group. If you do this correctly you will go through each of those elements and find that it does not generate the entire group, leading you to conclude that $\mathbb Z_2\times \mathbb Z_2$ is not cyclic.

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Whenever you have questions like this, it's always good to try a few small examples. The easiest cyclic group to think about is $\mathbb{Z}/2$, so it's natural to look at $\mathbb{Z}/2 \oplus \mathbb{Z}/2$. This group only has four elements, so you can easily see that this isn't cyclic just by looking at all the elements.

Then you might look at a group like $\mathbb{Z}/2 \oplus \mathbb{Z}/3$. This group has six elements, and without too much trouble you can find a generator for this group, and it is therefore isomorphic to $\mathbb{Z}/6$. You might try a few more examples, and see if you can conjecture what properties $m$ and $n$ must satisfy so that $\mathbb{Z}/m \oplus \mathbb{Z}/n$ is cyclic.

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Thanks for your help. –  Faye Feb 11 '13 at 7:10

Recall the Klein 4-group: This is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$, the direct product of the cyclic group $\mathbb{Z}_2$ with itself.

If you recall, although this group is abelian, it is not cyclic: there is no one element that generates all of $\mathbb{Z}_2 \times \mathbb{Z}_2$. It has three proper subgroups of order $2$ (in which the single non-identity element is its own inverse) and the trivial group, which you know contains only the identity.

Any direct product of $\mathbb{Z}_m$ and $\mathbb{Z}_n$ is non-cyclic, UNLESS $m, n$ are coprime: UNLESS, $\gcd(m,n) = 1$. When $\gcd(m, n) = 1$, then $\mathbb{Z}_m \times \mathbb{Z}_n \cong \mathbb{Z}_{mn}$, which IS cyclic.

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So, in some cases, the direct product of two cyclic groups is NOT cyclic, and in other cases, it IS cyclic. But to disprove your statement, you need only find ONE counterexample for which the statement is not true: and the direct product of $\mathbb{Z}_2 \times \mathbb{Z}_2$ is sufficient for that counterexample. –  amWhy Feb 11 '13 at 3:23
You gave the OP good example. Of course, he/she could find more. + –  Babak S. Feb 11 '13 at 6:06
@amWhy: Thanks for the feedback. –  Faye Feb 11 '13 at 7:11